Power: Definition, Concept And Example. Energy, Work Output.

Written on 19.07 by sianturikomkom

In daily life, it is interesting, not only to know the work that can be done, but also the speed or rate the work is done.

The work a person can do is limited not only by the total energy output, but also by the speed that the person is able to transform that energy into an specific work output.
Power can be defined either as the rate at which work is done, or, as the speed of energy transfer per unit time.
In the International System, power is expressed as
Joules per second, unit that receives the name
Watt (W), 1 W = 1J/s.
When we say a small bulb wastes 60 watts, we are saying it transforms 60 Joules of electric energy every second into luminous energy and thermal energy.
For large power measurements we use the Horse Power, abbreviated as Hp, equivalent to 746 Watts.
Sometimes it is convenient to express power in terms of the net force F applied to an object and its velocity.

P = W/t, and as W = Fd and v = d/t where d represents displacement, we get,
P = Fv, this is, force times velocity.

Example 1.
Calculate the power a 1200 kg automobile requires for the following situations:

a) The automobile moves up a long hill inclined 8º with a constant velocity of 12 m/s.
b) The automobile accelerates from 14 m/s to 18 m/s in 10 s to pass another vehicle, on an horizontal road. Suppose a constant friction force for both situations equal to Fr = 500 N.

F denotes the force generated by the engine.
SOLUTION.
a) At constant velocity the acceleration is zero, so we can write:

F = Fr+mgsen
F = 500 N+1200 kg·9,8 m/s² ·sen8º = 2.137 N

Using P = Fv, results P = 2.137N·12m/s = 25644 Watts, and when expressed in Hp, it is 34,3 Hp.
b) the acceleration is (18m/s-14m/s)/10s = 0,4 m/s².
By the Second Newton's Law, the sum of all the external forces must be equal to ma, mass times acceleration.

F - Fr = ma
F = 1200kg·0,4m/s² + 500N = 980 N

The necessary power to reach 18 m/s and pass the other vehicle is

P = Fv = 980N·18m/s = 17.640 Watts or 23,6 Hp.

Example 2.
An elevator whose mass is 1,195 kg can carry passengers with up to 935 kg mass. A 4,255 N constant friction force is opposite to the upward movement, as it is shown in the free body diagram below. a) What is the minimum power required by the engine elevator to make sure to lift the elevator at a constant speed of 4.5 m/s? b) What is the required power when the elevator is designed to develop an upward acceleration of 1.75 m/s² and the speed is 4.5 m/s?.
Solution:
a) The free body diagram indicates the engine must develop a force T to rise the elevator. f represents the constant friction force and mg, the elevator weight.

Constant speed means zero acceleration, a = 0.
Second Newton' Law states:
Where m is the total mass equal to 2,130 kg (elevator plus passengers). Then
T = f + mg
T = 4.255x10³ N + (2.13x10³kg)(9.8m/s²) T = 2.51x10⁴N
The power P is:
P = Tv = (2.51x10⁴ N)(4.5 m/s) = 11.3x10⁴ W
b) Now the engine must do additional work for accelerating the elevator. The only change in the setting of the problem is that a 0
F_y = T - f - mg = ma
T = m(a + g) + f
= (2.13x10³kg)(1.75 + 9.8)m/s² + 4,255 N = = 2.89x10⁴ N
The necessary power P is:
P = (2.89x10⁴ N)(4.5 m/s) = 13.0x10⁴ Watts

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